THERMODYNAMICS
What are the driving forces for the universe. Why do somethings occur without any
external assistance? There are countless reactions that happen spontaneously and
just as many that do not. So what governs spontaneity? Often your general chemical
knowledge is enough to evaluate a chemical reaction regarding the liklihood of
spontaneity.
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If a process is spontaneous, the reverse proccess is nonspontaneous.
Both are possible but spontaneous processes occur naturally.
We know that having a negative ∆H is a favorable sign for spontaneity. The universe
seeks the lowest energy. It also seeks the greatest entropy, or disorder. We
symbolize entropy with the letter S. If we evaluate a ∆S for a process where the
entropy increases, the sign for ∆S is positive. Examples of increases in entropy are:
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solids giving rise to liquids giving rise to gases
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numbers of gas molecules increase (prods - reacts)
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temperature increases
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dissolving processes
Entropy is based on two measurable quantities, heat and temperature. If something is
at a low temperature, a small addition of heat will have a great change in disorder.
Conversely, if something is at a very high temperature then it is already well disordered
and a small addition of heat will have little affect. ∆S = qrev./T (rev. stands for a
reversible process).
The Second Law of Thermodynamics states that the entropy of the universe is
increasing: ∆Suniv = ∆Ssys + ∆Ssurr > 0
If it were easy to measure the interactions of system and surroundings we could use the
second law to determine spontaneity. Since it is not, we develop a new criterion
specific to the system. This is called the Gibbs free energy and is defined as ∆G = ∆H T∆S.
• If ∆G is negative the process is spontaneous
• If ∆G is positive, the process is nonspontaneous
• If ∆G is zero, the process is at equilibrium
You should work at identifying qualitatively the sign of ∆G by examination of the
chemical reaction. For a relatively simple example, the equation
2N —> N2 shows a reduction in numbers of moles of gas so ∆S must be negative.
Secondly, the reaction has only bond formation which would be exothermic and thus ∆H
should be negative. Since both values are negative, ∆G must be negative (reaction
spontaneous) only at low temperatures. You can verify by examining the free energy
equation or by common sense (in this case, anyway). If the temperature is very high
the kinetic energy of the atoms will prevent bond formation!
Earlier, it was noted that at equilibrium, ∆G is zero. If this is the case, then the free
energy equation can be rearranged to: ∆S° = ∆H°/T. If you consider the phase change
for liquid water at 100.0°C to water vapor the ∆H is 44.0 kJ and T = 373.15. ∆S =
+117.9 J/K. The superscript ° means the transition occured at 1 atmosphere of
pressure.
For most liquids, the entropy of vaporization is approximately 88 J/mol K. This
generalization is known as Trouton's Rule. Since most gases occupy roughly 1000x the
liquid volume, the ∆Ss should be comparable. For liquids with strong intermolecular
forces (water) the value is too low. For weak intermolecular forces, the 88 J/mol K is
too high. Estimate the boiling point of Br2. How does this compare to the accepted
value?
The Third Law of Thermodynamics states: The entropy of a pure perfect crystal at 0 K
is zero. This is related to the quest to establish absolute entropies. Changes in entropy
during phase changes can be calculated with ∆S° = ∆H°/T as above. Other entropy
changes are derived from how specific heat changes with temperature. Ultimately, the
useful outcome is that one can calculate ∆S values from tabulated data in Appendix D.
∆S°rxn = ∑Sprods - ∑Sreact
• In general and for qualitative analysis, the more complex the molecules
(more atoms) the greater the molar entropies.
The standard free energy change is related to enthalpy and thus absolute values
cannot be established. But free energy changes can be determined. Like the creation
of ∆H°f, we can arbitrarily assign 0 free energy to the elements in their standard states
at 1 atm. pressure. We can then find
∆G°rxn = ∑∆G°prods - ∑∆G°react
Note that for this relationship, Hess's Law will work, if you reverse a reaction the value
of ∆G changes sign, and it is an extensive property. Please also note that this
particular equation is good only at 298.15°C!
Note that Example 20-6 shows that standard free energy can be calculated two ways.